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          <div class="post-toc animated"><ol class="nav"><li class="nav-item nav-level-1"><a class="nav-link" href="#%E7%BE%A4%E8%AE%BA%E7%AE%80%E4%BB%8B"><span class="nav-number">1.</span> <span class="nav-text">群论简介</span></a><ol class="nav-child"><li class="nav-item nav-level-2"><a class="nav-link" href="#%E5%AE%9A%E4%B9%89"><span class="nav-number">1.1.</span> <span class="nav-text">定义</span></a></li><li class="nav-item nav-level-2"><a class="nav-link" href="#%E5%87%A0%E4%B8%AA%E6%A0%97%E5%AD%90"><span class="nav-number">1.2.</span> <span class="nav-text">几个栗子</span></a></li><li class="nav-item nav-level-2"><a class="nav-link" href="#%E4%B8%80%E7%B1%BB%E7%89%B9%E6%AE%8A%E7%9A%84%E7%BE%A4"><span class="nav-number">1.3.</span> <span class="nav-text">一类特殊的群</span></a></li></ol></li><li class="nav-item nav-level-1"><a class="nav-link" href="#%E7%BD%AE%E6%8D%A2%E7%BE%A4"><span class="nav-number">2.</span> <span class="nav-text">置换群</span></a><ol class="nav-child"><li class="nav-item nav-level-2"><a class="nav-link" href="#%E4%B8%8D%E4%B8%A5%E8%B0%A8%E5%AE%9A%E4%B9%89"><span class="nav-number">2.1.</span> <span class="nav-text">不严谨定义</span></a></li><li class="nav-item nav-level-2"><a class="nav-link" href="#burnside-%E5%BC%95%E7%90%86"><span class="nav-number">2.2.</span> <span class="nav-text">burnside 引理</span></a><ol class="nav-child"><li class="nav-item nav-level-3"><a class="nav-link" href="#%E7%AE%80%E5%8C%96%E5%AE%9A%E4%B9%89"><span class="nav-number">2.2.1.</span> <span class="nav-text">简化定义</span></a></li><li class="nav-item nav-level-3"><a class="nav-link" href="#%E5%AE%8C%E6%95%B4%E5%AE%9A%E4%B9%89"><span class="nav-number">2.2.2.</span> <span class="nav-text">完整定义</span></a></li><li class="nav-item nav-level-3"><a class="nav-link" href="#%E9%80%82%E7%94%A8%E4%BA%8E%E7%AB%8B%E4%BD%93%E5%9B%BE%E5%BD%A2%E7%9A%84-burnside-%E5%BC%95%E7%90%86"><span class="nav-number">2.2.3.</span> <span class="nav-text">适用于立体图形的 burnside 引理</span></a><ol class="nav-child"><li class="nav-item nav-level-4"><a class="nav-link" href="#%E7%AB%8B%E6%96%B9%E4%BD%93%E9%9D%A2%E6%9F%93%E8%89%B2"><span class="nav-number">2.2.3.1.</span> <span class="nav-text">立方体面染色</span></a></li><li class="nav-item nav-level-4"><a class="nav-link" href="#%E6%AD%A3%E5%8D%81%E4%BA%8C%E9%9D%A2%E4%BD%93%E6%9F%93%E8%89%B2"><span class="nav-number">2.2.3.2.</span> <span class="nav-text">正十二面体染色</span></a></li><li class="nav-item nav-level-4"><a class="nav-link" href="#%E8%B6%B3%E7%90%83%E7%9A%84%E7%BD%AE%E6%8D%A2%E7%BE%A4"><span class="nav-number">2.2.3.3.</span> <span class="nav-text">足球的置换群</span></a></li></ol></li></ol></li></ol></li><li class="nav-item nav-level-1"><a class="nav-link" href="#%E4%BE%8B%E9%A2%98%EF%BC%9A"><span class="nav-number">3.</span> <span class="nav-text">例题：</span></a></li><li class="nav-item nav-level-1"><a class="nav-link" href="#%E9%A2%98%E5%A4%96%E8%AF%9D"><span class="nav-number">4.</span> <span class="nav-text">题外话</span></a></li></ol></div>
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          「学习总结」群 置换群
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        <p>群，一种特殊的代数结构，满足封闭性、结合律、存在幺元和对于每个元素存在逆元的四种性质。</p>
<p>置换群，一种特殊的群，其中的元素描述了一种交换操作。</p>
<a id="more"></a>
<h1 id="群论简介"><a href="#群论简介" class="headerlink" title="群论简介"></a>群论简介</h1><h2 id="定义"><a href="#定义" class="headerlink" title="定义"></a>定义</h2><p>称 集合 $S$ 和 $S$ 上运算 $\cdot$，共同构成得代数结构记作 $(S, \cdot)$，为一个群，当且仅当其满足如下性质：</p>
<ul>
<li>$S \not = \varnothing$</li>
<li>封闭性：$\forall a, b\in S, a\cdot b \in S$</li>
<li>结合律：$\forall a, b, c\in S, a \cdot (b \cdot c)= (a \cdot b) \cdot c$</li>
<li>幺元：$\exists e \in S, \forall a \in S, e \cdot a = a \cdot e= a$</li>
<li>逆元：$\forall a \in S, \exists b \in S, a \cdot b = e$</li>
</ul>
<h2 id="几个栗子"><a href="#几个栗子" class="headerlink" title="几个栗子"></a>几个栗子</h2><ul>
<li>实数集与实数间的乘法 $(\mathbb{R}, \times)$ 不是群，元素 $0 \in \mathbb{R}$，但是 $0$ 不存在逆元。</li>
<li>$(\mathbb{R’}, \times)$ 是一个群，幺元为 $1$。 其中 $\mathbb{R’} = \{\ x\ |\ x\in \mathbb{R}\ \land \ x \not = 0\ \}$</li>
<li>$(\mathbb{Z}, +)$ 是一个群，幺元是 $0$。</li>
<li>$(\mathbb{R}, +)$ 是一个群，幺元是 $0$。</li>
<li>$(\mathbb{P}, \times)$ 是一个群，幺元是 $1$ ，其中 $P$ 为某个质数的剩余系。</li>
</ul>
<h2 id="一类特殊的群"><a href="#一类特殊的群" class="headerlink" title="一类特殊的群"></a>一类特殊的群</h2><ul>
<li><strong>阿贝尔群</strong>：除了满足群的性质，还需要满足 $a \cdot b = b \cdot a$ 即 <strong>交换律</strong>。</li>
</ul>
<h1 id="置换群"><a href="#置换群" class="headerlink" title="置换群"></a>置换群</h1><h2 id="不严谨定义"><a href="#不严谨定义" class="headerlink" title="不严谨定义"></a>不严谨定义</h2><p>把 $1$ ~ $n$ 个对象 进行交换操作的群，置换群中集合的元素描述了一种交换操作。 对应的，置换群的运算就是分别执行两种交换操作。</p>
<div class="note info"><p>一种描述交换操作（置换）的记号。</p>
<p><strong>举个栗子：</strong><br>给执行交换操作前的元素依次编号为 $1, 2, 3, 4$（对编号后，编号为 $i$ 元素简称元素 i） ，交换操作后变为 $3, 1, 2, 4$。</p>
<p>可以描述成 $(1, 2, 3)(4)$，读作：将元素 $1$ 换到位置 $2$，将元素 $2$，换到位置 $3$，将元素 $3$ 换至位置 $1$，将元素 $4$ 换至位置 $4$。</p>
<p><strong>可以考虑成：</strong><br>一张包含 $n$ 个点的图，然后使 点$i$ 与 点$a_i$ 有边相连。对于每个连通块来说，图的形态一定是一个环，从环上一个点开始按照任意方向遍历图，得到的遍历顺序就是上面描述中的一个括号内的内容。联通块的个数就是上面描述中的括号数量，其实这个可以被称为轨道数。</p>
<p><strong>交换群的栗子</strong>：<br>一个对 $3$ 个元素的交换群：$\{e,(1,2)(3),(1,3)(2),(1,2,3),(1,3,2),(2,3)(1)\}$ </p>
<p>$(1)(2) \cdots (n)$ 简记为 $e$ 即 对原元素不做交换。</p>
</div>

<h2 id="burnside-引理"><a href="#burnside-引理" class="headerlink" title="burnside 引理"></a>burnside 引理</h2><details class="note danger"><summary><p>QAQ</p>
</summary>
<p><img src="/post_images/qun_0.png"><br><img src="/post_images/qun_1.png"></p>

</details>
<h3 id="简化定义"><a href="#简化定义" class="headerlink" title="简化定义"></a>简化定义</h3><p>设要对 $n$ 个元素用 $m$ 种颜色染色，对应置换群为 $S$，在该置换群下任意一种得到的相同方案算同一种方案。求本质不同的染色方案数。</p>
<p>根据 burnside 引理，其答案为。<br>$$<br>\begin{equation}<br>   ANS = \frac{1}{|S|}\sum_{s \in S}\limits{m^{\eta(s)}} \label{bns}<br>\end{equation}<br>$$<br>其中 $\eta(s)$ 为置换方案 $s$ 的轨道数。 这是关于 burnside 引理的一个<strong>简化版</strong>定义。</p>
<div class="note info"><p><strong>一个栗题</strong>：<br>$n$ 个排成一圈的点用 $m$ 种颜色染色，问方案数（旋转前后的方案为一种方案）.</p>
</div>
<details class="note success"><summary><p>solution</p>
</summary>
<p>显然需要解决两个问题：</p>
<ul>
<li>置换群的构造.</li>
<li>置换群元素轨道数目.</li>
</ul>
<p>考虑这个置换群的置换集合可以是什么： (假设 $n$ 为 $6$).<br>$$<br>\begin{equation*}<br>S = \begin{Bmatrix}<br>(1)(2)(3)(4)(5)(6), \\<br>(1, 2, 3, 4, 5, 6), \\<br>(1, 3, 5) (2, 4, 6) \\<br>(1, 4) (2, 5) (3, 6) \\<br>(1, 5, 3) (2, 6, 4) \\<br>(1, 6, 5, 4, 3, 2) \\<br>\end{Bmatrix}<br>\end{equation*}<br>$$<br>分别对应转动的位置个数为 $0$ ~ $n - 1$。稍微模拟一下这个过程，显然移动 $k$ 个位置的方案，轨道数为 $\operatorname{gcd}(k, n)$. 然后就一般化了.</p>

</details>

<h3 id="完整定义"><a href="#完整定义" class="headerlink" title="完整定义"></a>完整定义</h3><div class="note danger"><p>设 $A$ 和 $B$ 为有限集合，$X = B^A$ 表示从 $A$ 到 $B$ 的映射。$G$ 是 $A$ 上置换群，$X / G$ 表示 $G$ 作用在 $X$ 上产生的所有等价类集合，若存在两种映射经过 $G$ 的置换作用后是相同的，则将两种映射视为同一种。<br>burnside 引理指出：<br>$$<br>\begin{equation}<br>|X / G| = \frac{1}{|G|}\sum_{g \in G} \limits{|X^g|} \label{burnside}<br>\end{equation}<br>$$<br>其中 $X^g$ 表示在映射集合 $X$ 中有多少元素可以经过 $g$ 的变化，变成一样的。</p>
</div>
<p>考虑与之前定义 $(\ref{bns})$ 的联系：</p>
<ul>
<li>一种染色方案可以看成 元素 和 颜色 的映射。</li>
<li>对于置换 $s$ ，如果要保证其作用前后映射方案相同，必须要保证其同轨道内的元素被映射到了同一元素，即 染成了相同的颜色。<ul>
<li>那么考虑置换 $s$ 的每一条轨道 $\eta$，每一条轨道颜色相同的方案数就是 $m^{\eta(s)}$ </li>
</ul>
</li>
<li>这与 $(\ref{burnside})$ 的定义是相符的。</li>
</ul>
<div class="note info"><p><strong>一道例题</strong><br>众所周知，小葱同学擅长计算，尤其擅长计算组合数，但这个题和组合数没什么关系。</p>
<p>现在有一个迷宫，这个迷宫是由若干个正 $n(=4,6)$ 边形组成的k层迷宫。如果 $k=1$ ，那么该迷宫就由单独一个正 $n$ 边形组成；如果 $k&gt;1$，则在 $k−1$ 层的基础上，沿着所有最外层的边增加一个正 $n$ 边形，新增加的正 $n$ 边形若有重叠，则保留其中一个即可。具体可以参考下图： </p>
<p><img src="/post_images/qun_2.png"></p>
<p>现在为了打破迷宫的结界，你需要在迷宫的某些边上开一扇门。你总共需要开 $r$ 扇门，每条边最多打开一扇门。但是如果两种开门的方案通过旋转相同，那么视为同一种方案。以及由于是死亡迷宫，所以死了也是可以的，所以你并不需要保证你开门的方案能够让你走出去。求总共的方案数。</p>
</div>

<details class="note success"><summary><p>solution</p>
</summary>
<p>现在可以简化一下描述一种操作的语言，burnside 引理 不关心置换的具体对象，只关心置换的大小和出现次数。可以用简写 $(a)^b$ 表示：有 $b$ 个长度为 $a$ 的轨道。<br>先特殊考虑一下 $k = 2$ 的情况，一共有 $16$ 条边。可以分为转 $0^o,90^o,180^o, 270^o$.</p>
<ul>
<li>转 $0^o$ 的置换，可以描述为 $(1)^{16}$.</li>
<li>转 $90^o$ 的置换，可以描述为 $(4)^4$.</li>
<li>转 $180^o$ 的置换，可以描述为 $(2)^8$.</li>
<li>转 $270^o$ 的置换，可以描述为 $(4)^4$.</li>
</ul>
<p>特殊化一下，不难发现就是：（一共有 $4k^2$ 条边） </p>
<ul>
<li>转 $0^o$ 的置换，可以描述为 $(1)^{4k^2}$.</li>
<li>转 $90^o$ 的置换，可以描述为 $(4)^{k^2}$.</li>
<li>转 $180^o$ 的置换，可以描述为 $(2)^{2k^2}$.</li>
<li>转 $270^o$ 的置换，可以描述为 $(4)^{k^2}$.</li>
</ul>
<p>根据 $X^g$ 的定义——表示在映射集合 $X$ 中有多少元素可以经过 $g$ 的变化，变成一样的。我们始终要保证对于一种置换，同一轨道的元素完全相同.<br>所以对于第一种置换，贡献就是在轨道数中选出 $r$ 个让他们都变成 “有门存在” 的状态，即 $\dbinom{4k^2}{r}$.<br>其余的同理，答案就是。<br>$$<br>\begin{equation}<br>   \frac{1}{4}\sum_{g\in G}\limits{\dbinom{\eta(g)}{\frac{r}{t}}}<br>\end{equation}<br>$$<br>其中 $t$ 为 轨道 $g$ 的循环长度。 当 $\frac{r}{t}$ 不是整数的时候，贡献为 $0$ ，因为 不存在某种染色方法，使得这种置换中同轨道的元素颜色相同.</p>

</details>

<h3 id="适用于立体图形的-burnside-引理"><a href="#适用于立体图形的-burnside-引理" class="headerlink" title="适用于立体图形的 burnside 引理"></a>适用于立体图形的 burnside 引理</h3><p>包括对正 $n$ 面体和足球的 面，点或棱染色。</p>
<h4 id="立方体面染色"><a href="#立方体面染色" class="headerlink" title="立方体面染色"></a>立方体面染色</h4><p>首先讨论置换群：<br>考虑正方体有 $12$ 条棱，$6$ 个面，$8$ 个点。</p>
<table>
<thead>
<tr>
<th>旋转轴类型</th>
<th>角度</th>
<th>数量</th>
<th>置换</th>
</tr>
</thead>
<tbody><tr>
<td>不转</td>
<td>$0^o$</td>
<td>1</td>
<td>$(1)^6$</td>
</tr>
<tr>
<td>面面</td>
<td>$90^o$</td>
<td>3</td>
<td>$(1)^2(4)^1$</td>
</tr>
<tr>
<td>面面</td>
<td>$180^o$</td>
<td>3</td>
<td>$(1)^2(2)^2$</td>
</tr>
<tr>
<td>面面</td>
<td>$270^o$</td>
<td>3</td>
<td>$(1)^2 (4)^1$</td>
</tr>
<tr>
<td>棱棱</td>
<td>$180^o$</td>
<td>6</td>
<td>$(2)^3$</td>
</tr>
<tr>
<td>点点</td>
<td>$120^o$</td>
<td>4</td>
<td>$(3)^2$</td>
</tr>
<tr>
<td>点点</td>
<td>$240^o$</td>
<td>4</td>
<td>$(3)^2$</td>
</tr>
</tbody></table>
<p>关于角度的确定：观察旋转轴图形周围的形状。数量为总数的一半。</p>
<h4 id="正十二面体染色"><a href="#正十二面体染色" class="headerlink" title="正十二面体染色"></a>正十二面体染色</h4><p>$20$ 个点，$12$ 个面，$30$ 条棱，面为五边形。</p>
<div class="note warning"><p>循环节：即对于一种置换方案的轨道长度。简单理解为转多少次能够转到原来的位置。</p>
</div>
<table>
<thead>
<tr>
<th>旋转轴类型</th>
<th>角度</th>
<th>循环节</th>
<th>数量</th>
<th>面染色置换群</th>
<th>边染色置换群</th>
<th>点染色置换群</th>
</tr>
</thead>
<tbody><tr>
<td>不转</td>
<td>$0^o$</td>
<td>1</td>
<td>1</td>
<td>$(1)^{12}$</td>
<td>$(1)^{30}$</td>
<td>$(1)^{20}$</td>
</tr>
<tr>
<td>面面轴</td>
<td>$72^o$</td>
<td>5</td>
<td>6</td>
<td>$(1)^2(5)^2$</td>
<td>$(5)^6$</td>
<td>$(5)^4$</td>
</tr>
<tr>
<td>面面轴</td>
<td>$144^o$</td>
<td>5</td>
<td>6</td>
<td>$(1)^2(5)^2$</td>
<td>$(5)^6$</td>
<td>$(5)^4$</td>
</tr>
<tr>
<td>面面轴</td>
<td>$216^o$</td>
<td>5</td>
<td>6</td>
<td>$(1)^2(5)^2$</td>
<td>$(5)^6$</td>
<td>$(5)^4$</td>
</tr>
<tr>
<td>点点轴</td>
<td>$120^o$</td>
<td>3</td>
<td>10</td>
<td>$(3)^4$</td>
<td>$(3)^{10}$</td>
<td>$(1)^2(3)^6$</td>
</tr>
<tr>
<td>点点轴</td>
<td>$240^o$</td>
<td>3</td>
<td>10</td>
<td>$(3)^4$</td>
<td>$(3)^{10}$</td>
<td>$(1)^2(3)^6$</td>
</tr>
<tr>
<td>棱棱轴</td>
<td>$180^o$</td>
<td>2</td>
<td>15</td>
<td>$(2)^6$</td>
<td>$(1)^2(2)^{14}$</td>
<td>$(2)^{10}$</td>
</tr>
</tbody></table>
<p>以某个对象为旋转轴，其置换群中必然有两个长度为 $1$ 的轨道，作为旋转轴的对象，旋转前后应该是重叠的。<br>然后剩下的根据轨道循环节直接算出即可。 可以发现这是一个无脑的工作</p>
<h4 id="足球的置换群"><a href="#足球的置换群" class="headerlink" title="足球的置换群"></a>足球的置换群</h4><details class="note warning"><summary><p>qaq</p>
</summary>
<p>实在不想写了……<br>足球的点数不好求，但是可以根据立体图形外角和公式求出：<br>立体图形外角和为 $720^o$ 。 即 $\text{点数} \times \text{外角度} = 720^o$.</p>

</details>


<h1 id="例题："><a href="#例题：" class="headerlink" title="例题："></a>例题：</h1><p>待填。</p>
<h1 id="题外话"><a href="#题外话" class="headerlink" title="题外话"></a>题外话</h1><details class="note danger"><summary><p>呵呵</p>
</summary>
<p><img src="/post_images/qun_3.png"></p>

</details>
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